SOLUTION: let cos A= 1/5^(1/2) with A in Q 4. find tan 2A

let cos(A)= 1/5^(1/2) with A in Q 4. find tan(2A)

------------------------

The 1/2 power is the same as the square root, so
            _ 
cos(A) = 1/Ö5

The formula for tan(2A) is

            2·tan(A)
tan(2A) = -------------
           1 - tan²(A)
                                           sin(A) 
so we need tan(A).  We know that tan(A) = -------- 
                                           cos(A)

and we have cos(A), so we need to get sin(A).

We know that sin²(A) = 1 - cos²(A), so substituting
   _
1/Ö5 for cos(A),
                  _
sin²(A) = 1 - (1/Ö5)² = 1 - 1/5 = 4/5

Taking square roots:
           _____        _
sin(A) = ±Ö(4/5) = ± 2/Ö5

Since we know that A is in Q 4, and since the
sine is negative in the 4th quadrant,
we must take the negative sign, so
             _
sin(A) = -2/Ö5              _
             sin(A)     -2/Ö5      
So tan(A) = -------- = -------- 
             cos(A)      1/Ö5        
                            _
Multiply top and bottom by Ö5 and you get:

tan(A) = -2

Now we plug that in:

            2·tan(A)         2(-2)         -4      -4      4
tan(2A) = ------------- = ----------- = ------- = ----- = ---
           1 - tan²(A)     1 - (-2)²     1 - 4     -3      3

Edwin